Question

A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB (= a) subtends an angle of ${60}^{\circ }$ at the foot of the tower, and the angle of elevation of the top of the tower from A or B is ${30}^{\circ }$. The height of the tower is

• A. $\frac{2a}{\sqrt{3}}$
• B. $2\sqrt{3}a$
• C. $\frac{a}{\sqrt{3}}$
• D. $a\sqrt{3}$

Answer 1

The correct option is C $\frac{a}{\sqrt{3}}$

Let us consider the height of the tower as $h$
We know that, $h=AC\tan {30}^{\circ }=BC\tan {30}^{\circ }$ ...(1)


We know that $AC=BC$ [$\because$ radius of circle]

Hence, we can say that $△ABC$ is an isosceles triangle with $AC=BC$.
So, $\angle ABC=\angle BAC$ ...(2)

But $\angle ACB={60}^{\circ }$ [given]
$\Rightarrow \angle ABC+\angle BAC+\angle ACB={180}^{\circ }$
$\Rightarrow \angle ABC+\angle ABC+{60}^{\circ }={180}^{\circ }$ [from (2)]
$\Rightarrow 2\times \angle ABC={120}^{\circ }$
$\therefore \angle ABC=\angle BAC={60}^{\circ }$

Hence $△ABC$ is an equilateral triangle.

$\therefore AB=BC=CA=a$ ...(3)


$\therefore h=a\tan {30}^{\circ }$ [form equation (1)]
$=\frac{a}{\sqrt{3}}$